What is the integral of (20sin(10^{11}t-4y)) ?

The integral of (20sin(10^{11}t-4y)) is -1/5000000000 cos(10^{11}t-4y)+C

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integral of (20sin(10^{11}t-4y))

\int (20 sin (1 0^{11} t - 4 y)) d t

- \frac{1}{5000000000} cos (1 0^{11} t - 4 y) + C

Solution steps

\int 20 sin (1 0^{11} t - 4 y) d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 20 \cdot \int sin (1 0^{11} t - 4 y) d t

Apply u-substitution

= 20 \cdot \int sin (u) \frac{1}{1 0 ^{11}} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 20 \cdot \frac{1}{1 0 ^{11}} \cdot \int sin (u) d u

Use the common integral:

\int sin (u) d u = - cos (u)

= 20 \cdot \frac{1}{1 0 ^{11}} (- cos (u))

Substitute back

u = 1 0^{11} t - 4 y

= 20 \cdot \frac{1}{1 0 ^{11}} (- cos (1 0^{11} t - 4 y))

Simplify

20 \cdot \frac{1}{1 0 ^{11}} (- cos (1 0^{11} t - 4 y)) : - \frac{1}{5000000000} cos (1 0^{11} t - 4 y)

= - \frac{1}{5000000000} cos (1 0^{11} t - 4 y)

Add a constant to the solution

= - \frac{1}{5000000000} cos (1 0^{11} t - 4 y) + C

What is the integral of (20sin(10^{11}t-4y)) ?
The integral of (20sin(10^{11}t-4y)) is -1/5000000000 cos(10^{11}t-4y)+C